Continuity Does Not Preserve Hausdorffness
Hausdorffness is a condition on how the open sets of a space separate pairs of points. Continuity, in contrast, asks whether preimages of open sets are open. Because these conditions inspect different topological data, continuity alone need not preserve Hausdorffness: a continuous surjection can start with a Hausdorff space and end with a non-Hausdorff one.
We will construct such maps using the indiscrete topology. It has so few open sets that continuity into it imposes almost no restriction on a map. The examples will also show why checking whether every fiber is open is not the same as checking continuity.
In the Hausdorff definition, the sets \(U\) and \(V\) are allowed to depend on the chosen pair \(x,y\). In the continuity definition, the sets tested are the open sets of the codomain—not every subset of the codomain. This distinction is decisive when the codomain is indiscrete: there are only two sets to test.
Why an Indiscrete Space with Two Points Is Not Hausdorff
Let \(Y=\{u,v\}\), where \(u\ne v\), with the indiscrete topology \(\{\varnothing,Y\}\). The only open set containing \(u\) is \(Y\), and the only open set containing \(v\) is also \(Y\). Thus any open neighborhoods chosen for \(u\) and \(v\) intersect in \(Y\), so they cannot be disjoint. Therefore \(Y\) is not Hausdorff.
More generally, an indiscrete space with at least two points is not Hausdorff: for any two distinct points, the only open neighborhood of either point is the whole space. The one-point indiscrete space is Hausdorff, since there are no distinct points to separate. Thus the failure in our examples comes from the presence of distinct points, not merely from the word “indiscrete.”
Proof. The open sets of \(Y\) are exactly \(\varnothing\) and \(Y\). For any map \(f:X\to Y\), we have \(f^{-1}(\varnothing)=\varnothing\) and \(f^{-1}(Y)=X\). Both sets are open in \(X\), by the axioms of a topology. These are all the preimages required in the definition of continuity, so \(f\) is continuous. If \(Y\) has at least two points, choose distinct \(u,v\in Y\). Their only open neighborhoods are both \(Y\), and \(Y\cap Y=Y\ne\varnothing\). Hence \(Y\) is not Hausdorff. \(\square\)
The proposition gives a direct way to build counterexamples. The source can have a topology that separates points very effectively, while the indiscrete codomain has too few open sets to separate any distinct points. Continuity checks only the codomain’s open sets and therefore does not force its points to become separable.
Worked Example: A Finite Hausdorff Source and an Indiscrete Target
Let \(X=\{0,1\}\) have the discrete topology \(\mathcal{P}(X)\), so every subset is open. Let \(Y=\{u,v\}\) have the indiscrete topology \(\{\varnothing,Y\}\), and define \(f:X\to Y\) by \(f(0)=u\) and \(f(1)=v\).
First, \(f\) is surjective: \(u=f(0)\) and \(v=f(1)\), so every point of \(Y\) is attained. The source \(X\) is Hausdorff. Indeed, for its two distinct points, \(\{0\}\) and \(\{1\}\) are open, contain \(0\) and \(1\) respectively, and have empty intersection.
To check continuity, test each open set in \(Y\). The preimages are \(f^{-1}(\varnothing)=\varnothing\) and \(f^{-1}(Y)=X\), both open in \(X\). Thus \(f\) is continuous. But the target is not Hausdorff: its only nonempty open set is \(Y\), so \(u\) and \(v\) have no disjoint open neighborhoods.
This map is even a bijection, but it is not a homeomorphism. A homeomorphism is a bijection that is continuous and whose inverse is continuous. Here \(f^{-1}(\{u\})=\{0\}\), but \(\{u\}\) is not open in \(Y\), so the inverse map does not take this open subset \(\{0\}\) of \(X\) to an open set of \(Y\). In particular, a continuous bijection need not preserve Hausdorffness in the direction considered here.
A Surjection from the Real Line
The finite example already proves the central point, but a map from the real line makes the role of fibers especially visible. Throughout the next examples, \(\mathbb{R}\) has its standard topology.
Worked Example: A Two-Value Map with One Nonopen Fiber
Let \(Y=\{u,v\}\) have the indiscrete topology, and define \(f:\mathbb{R}\to Y\) by \(f(x)=u\) when \(x<0\), and \(f(x)=v\) when \(x\ge 0\).
The map is surjective: \(f(-1)=u\), since \(-1<0\), and \(f(0)=v\), since \(0\ge 0\). Its fibers are \(f^{-1}(\{u\})=(-\infty,0)\) and \(f^{-1}(\{v\})=[0,\infty)\). The first fiber is open in the standard topology on \(\mathbb{R}\), while the second is not open: every open interval containing \(0\) also contains negative real numbers, so no open neighborhood of \(0\) is contained in \([0,\infty)\).
Despite that nonopen fiber, \(f\) is continuous. The open sets of the codomain are \(\varnothing\) and \(Y\), and their preimages are \(f^{-1}(\varnothing)=\varnothing\) and \(f^{-1}(Y)=\mathbb{R}\). Both are open in \(\mathbb{R}\). The singletons \(\{u\}\) and \(\{v\}\) are not open in \(Y\), so their preimages are not among the sets that continuity requires us to test.
The source is Hausdorff: given distinct real numbers \(x<y\), set \(r=(y-x)/3\). Then \(r>0\), and the open intervals \((x-r,x+r)\) and \((y-r,y+r)\) contain \(x\) and \(y\), respectively. They are disjoint because \(x+r<y-r\), which follows from \(2r=2(y-x)/3<y-x\). The target is not Hausdorff, as shown above. Hence this is a continuous surjection from a Hausdorff space onto a non-Hausdorff space.
This example corrects a tempting but incorrect shortcut: continuity does not require every fiber \(f^{-1}(\{y\})\) to be open. Such a requirement follows from continuity only when \(\{y\}\) is open in the codomain. The definition instead requires open preimages for open codomain sets, and in an indiscrete space the only such sets are the empty set and the whole space.
Worked Example: A Three-Value Surjection
Let \(Y=\{a,b,c\}\) have the indiscrete topology \(\{\varnothing,Y\}\). Define \(g:\mathbb{R}\to Y\) by \(g(x)=a\) for \(x<0\), \(g(x)=b\) for \(0\le x<1\), and \(g(x)=c\) for \(x\ge1\).
The three cases are disjoint and cover \(\mathbb{R}\): each real number either is negative, belongs to \([0,1)\), or is at least \(1\). The map is surjective because \(g(-1)=a\), \(g(0)=b\), and \(g(1)=c\), using the stated endpoint conventions. The fibers are \(g^{-1}(\{a\})=(-\infty,0)\), \(g^{-1}(\{b\})=[0,1)\), and \(g^{-1}(\{c\})=[1,\infty)\). The first is open, while the second and third are not open: every open interval around \(0\) contains negative numbers, and every open interval around \(1\) contains numbers less than \(1\). Thus neither \([0,1)\) nor \([1,\infty)\) is a neighborhood of each of its points contained in that fiber.
For continuity, there are again only two open sets in the codomain. Their preimages are \(g^{-1}(\varnothing)=\varnothing\) and \(g^{-1}(Y)=\mathbb{R}\), both open. Therefore \(g\) is continuous, although two of its fibers are not open. Finally, \(Y\) is not Hausdorff: for example, any open neighborhoods of \(a\) and \(b\) must both equal \(Y\), and their intersection is \(Y\), not the empty set.
What the Examples Establish
The construction is not limited to these particular formulas. The proposition proves that every map into an indiscrete space is continuous, regardless of the topology on the source. A surjection from a Hausdorff source is then possible whenever the source has enough points to map onto the chosen target. For example, given any Hausdorff space \(X\) with at least two points, choose \(x_0\in X\) and let \(Y=\{u,v\}\) be indiscrete. Define \(f(x_0)=u\) and \(f(x)=v\) for every \(x\ne x_0\). Since \(X\) has a point other than \(x_0\), the map is surjective; the proposition gives continuity, and \(Y\) is not Hausdorff.
The example does not say that every feature of a space is ignored by continuity. Rather, continuity is measured against the specified topology of the codomain. A codomain with few open sets makes the preimage condition weak. A Hausdorff source, by contrast, has enough open sets to separate its distinct points, but that source property does not impose a separation condition on the codomain through continuity alone.
Common pitfall. A wrong continuity check for the real-line example is: “Since \(f^{-1}(\{v\})=[0,\infty)\) is not open, \(f\) is not continuous.” The calculation of the fiber is correct, but the conclusion uses the wrong test set: \(\{v\}\) is not open in the indiscrete codomain. The correct check is \(f^{-1}(\varnothing)=\varnothing\) and \(f^{-1}(Y)=\mathbb{R}\), both open. It is also important not to claim that neither fiber is open: \(f^{-1}(\{u\})=(-\infty,0)\) is open, whereas \(f^{-1}(\{v\})=[0,\infty)\) is not.
The essential lesson is precise: Hausdorffness concerns disjoint neighborhoods of distinct points, while continuity concerns preimages of open sets. The indiscrete topology supplies no disjoint neighborhoods for distinct points and very few open sets for continuity to test. Consequently, continuity—and even continuous surjectivity—does not in general preserve Hausdorffness.
Check Your Understanding
For each question, keep the topology on both source and target in view.
- Why is a two-point set with the indiscrete topology not Hausdorff?
- Let \(X\) be any topological space and \(Y\) an indiscrete space. What are the possible preimages of open subsets of \(Y\), and why are they open?
- For the map \(f:\mathbb{R}\to\{u,v\}\) defined by \(f(x)=u\) for \(x<0\) and \(f(x)=v\) for \(x\ge0\), which fiber is not open? Why does this not contradict continuity?
- Can a continuous bijection from a Hausdorff space onto a non-Hausdorff space be a homeomorphism? Explain using the finite example.
- What condition on the size of an indiscrete target is needed for it to fail Hausdorffness?